Prove this identity: $ \tan(2x)-\sec(2x) =\tan(x-\pi/4)$ I've been having a time with this problem. I tried to start with the left side but I hit a dead end quick... I then tried the right side and had a little more luck but I've hit a block. I first used the tan difference identity then converted everything into sins and cosines. Then I used a difference identity on sin and cosine and then converted into exact values. $\dfrac{\pi}{4}$ for sine and cosine is $\dfrac{\sqrt{2}}{2}$. All the radicals cancelled out leaving me with $\dfrac{\sin x - \cos x }{\cos x + \sin x}$... I can't think anything to do with this... where did I go wrong?
Any answer is appreciated.
 A: You're almost there.  Dividing the numerator and denominator by $\cos x$ gives $$\frac{\sin x - \cos x}{\cos x + \sin x} = \frac{\tan x - 1}{1 + \tan x},$$ and then observe that $\tan \frac{\pi}{4} = 1$.  Then recall the tangent addition identity $$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}.$$  For what suitable choices of $\alpha$, $\beta$, does the identity match the expression you have?

Okay, so since there's some confusion about the direction in which you're going, and since you already basically have done the proof and just need to see it put all together, here it is:  $$\begin{align*} \tan(x - \tfrac{\pi}{4}) &= \frac{\tan x - \tan \frac{\pi}{4}}{1 + \tan x \tan \frac{\pi}{4}} \\ &= \frac{\tan x - 1}{1 + \tan x} \\ &= \frac{\sin x - \cos x}{\cos x + \sin x} \\ &= \frac{-(\sin x - \cos x)^2}{\cos^2 x - \sin^2 x} \\ &= \frac{- \sin^2 x + 2 \sin x \cos x - \cos^2 x}{\cos 2x} \\ &= \frac{-1 + \sin 2x}{\cos 2x} \\ &= \tan 2x - \sec 2x. \end{align*}$$  Going in the other direction, we might write $$\begin{align*} \tan 2x - \sec 2x &= \frac{\sin 2x}{\cos 2x} - \frac{1}{\cos 2x} \\ &= \frac{2 \sin x \cos x - 1}{\cos^2 x - \sin^2 x} \\ &= \frac{2 \sin x \cos x - \sin^2 x - \cos^2 x}{(\cos x + \sin x)(\cos x - \sin x)} \\ &= \frac{-(\cos x - \sin x)^2}{(\cos x + \sin x)(\cos x - \sin x)} \\ &= \frac{\sin x - \cos x}{\cos x + \sin x} \\ &= \frac{\tan x - 1}{1 + \tan x} \\ &= \frac{\tan x - \tan \frac{\pi}{4}}{1 + \tan x \tan \frac{\pi}{4}} \\ &= \tan(x - \tfrac{\pi}{4}).\end{align*}$$
A: $$\tan2x-\sec2x=\frac{\sin2x-1}{\cos2x}$$
Using  Weierstrass substitution, this becomes 
$$\frac{\dfrac{2t}{1+t^2}-1}{\dfrac{1-t^2}{1+t^2}}=-\frac{(1-t)^2}{1-t^2}$$ where $t=\tan x$
If $1-t\ne0\iff t\ne1$ i.e.,$\tan x\ne1$ this can be reduces to  $$-\frac{1-t}{1+t}=\frac{t-1}{t+1}=\frac{\tan x-\tan\dfrac\pi4}{1+\tan x\tan\dfrac\pi4}=\cdots$$
A: $$
\begin{align}
\tan(2x)-\sec(2x)
&=\frac{\sin(2x)-1}{\cos(2x)}\tag{1}\\[6pt]
&=\frac{\cos\left(2\left(\frac\pi4-x\right)\right)-1}{\sin\left(2\left(\frac\pi4-x\right)\right)}\tag{2}\\
&=\frac{-2\sin^2\left(\frac\pi4-x\right)}{2\sin\left(\frac\pi4-x\right)\cos\left(\frac\pi4-x\right)}\tag{3}\\[9pt]
&=-\tan\left(\frac\pi4-x\right)\tag{4}\\[18pt]
&=\tan\left(x-\frac\pi4\right)\tag{5}
\end{align}
$$
Explanation:
$(1)$: $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$ and $\sec(\theta)=\frac1{\cos(\theta)}$
$(2)$: $\sin\left(\frac\pi2-\theta\right)=\cos(\theta)$ and $\cos\left(\frac\pi2-\theta\right)=\sin(\theta)$
$(3)$: $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ and $\cos(2\theta)=1-2\sin^2(\theta)$
$(4)$: $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$
$(5)$: $\tan(-\theta)=-\tan(\theta)$

Starting where you left off, we can use the formulas for the sine and cosine of a difference to get
$$
\begin{align}
\frac{\sin(x)-\cos(x)}{\cos(x)+\sin(x)}
&=\frac{\sqrt2\sin\left(x-\frac\pi4\right)}{\sqrt2\cos\left(x-\frac\pi4\right)}\\
&=\tan\left(x-\frac\pi4\right)\tag{6}
\end{align}
$$
